Transformer Shunt Inductance Tuned Transformers
In the last lecture, we derived the transformer equations
where Im = magnetization current and It = transformer current.
An equivalent electrical circuit for such a nonideal transformer
can be constructed from these two equations
In particular, we have a shunt inductor that appears at the
primary terminals of an ideal transformer.
With the shunt inductance in the model, the high-pass nature of
a physical transformer is properly accounted for, since at DC the
primary terminals of T will be shorted.
Without the shunt L, the ideal transformer appears to transform
voltages and currents equally well for all frequency, which
cannot be true (by Faraday’s Law). Secondary Inductance
Equivalently, the shunt inductance can also have been
incorporated from the secondary.To do this, we begin again with
Now, from (6.10) Vs = Ns j??m
Substituting (3) into (2) leaves
The equivalent electrical circuit for (6.21) and this last
To check the directions for the current shown in this figure, we
can apply KCL at the secondary:
It = Im + Is
Is = It ? Im
The minus sign here agrees with (5).
To summarize this work so far, whether the magnetization
current effect is included on the primary side or the secondary
side of the transformer is immaterial: they are equivalent.
Actually, we can develop this latter secondary inductance
equivalent circuit simply from the impedance transformation
property of the ideal transformer! Begin with
We’ve seen previously that for an ideal transformer
Here Zp = j? Lp and Lp = Al Np2
which we can model simply as an ideal transformer with a shunt
Ls as shown on the previous page Tuned Transformers
Achieving impedance match between the various subsystems in
a multistage communications circuit is very important.
Otherwise, precious signal is needlessly wasted.
Transformers – specifically ideal transformers – can be used as
matching networks since, as we’ve already seen,
We can choose Np/Ns to change (or “transform”) Zp to a desired
value for matching.
Note that (6.19) is valid only for ideal transformers. One way to
negate the effects of the magnetization current Im in a practical
transformer (so that the ideal T equations apply) is to use a
We can adjust C to resonate out the effects of Lp at the desired
frequency of operation. That is, suppose the transformer is
designed to operate at f = f0. For an LC resonance at
f0 =1/ (2?under root LC ), then adjust C such that
Consequently, now at this operating frequency f0
and the equivalent circuit for this tuned transformer circuit
which is simply an ideal transformer. Very cool!
This resonant method is only a narrow-band solution, but it can
be extremely useful. Capacitive transformer tuning effectively
makes a band-pass filter from a high-pass filter. Examples
The two tuned transformers in the NorCal 40A are T2 (RF
Filter) and T3 (matching between RF Mixer and IF Filter). Let’s
consider both of these quickly once again in the light of our
expanded understanding of transformers
1. T3 (between RF Mixer and IF Filter): This transformer was
also considered in the previous lecture. T3 is used to transform
the output impedance from the RF Mixer to match the input
impedance of the IF Filter (200 ?).
From the data sheet for the SA602AN IC, the output impedance
is 2×1500 ? = 3000 ?. Using (7):
which is very close to the desired 200 ? for the IF Filter!
2. T2 (RF Filter): Consider once again the second order
Butterworth bandpass filter example we discussed earlier in
L1 and C1 are components that are soldered onto your PCB.
Where do the 86.2 nH and 6.0 nF components come from? As
we mentioned earlier in Lecture 12, they both come from T2!
To see this explicitly for T2, Lp = Al Np = 66 *10-9 H/turn2 12 66 nH, which is close to the 86.2 nH shown above that is needed for a second-order bandpass filter. What about the C? That comes from C2 and the impedance transforming properties of T2:
With Ns = 20, Np = 1 and Cs = 15 pF, then
which is exactly the value needed for the second order
Butterworth bandpass filter! Very cool.